Find the sample standard deviation s of the data: 55, 51, 17, 32, 19, 39, 60.
n = 7
s =
18.Descriptive statisticsHard
A quantity is normally distributed with mean 70 and standard deviation 20. What proportion P of values is below 82?
μ = 70σ = 20x = 82
P =
19.Normal distributionMedium
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. What fraction lies between 438 and 574 h? Give P as a decimal.
μ = 500σ = 40x1 = 438x2 = 574
P =
20.Normal distributionHard
The battery lifetimes are normally distributed with mean 500 h and standard deviation 40 h. Find the value x below which 97.5% of them lie (the 97.5th percentile).
μ = 500σ = 40P = 97.5%
x = h
3Solutions
1.Limits
Answer: L = −1.25
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
Multiply by the conjugate
f(x) = −5x/(√(4x2 − 5x) + 2x)
Divide by x
f(x) = −5/(√(4 + (−5)/x) + 2)
Let x → ∞
L = −5/(2 + 2) = −1.25
Answer
L = −1.25
2.Limits
Answer: L = 0.5
Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).
First application
(1·ex − 1)/(2x) → 0/0
Second application
1·ex/2
Substitute x = 0
L = 1·1/2 = 0.5
Answer
L = 0.5
3.Derivatives
Answer: f′(1) = 108
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Chain rule
f′(x) = 3·(4x − 1)2·4
Substitute
f′(1) = 3·32·4 = 108
Answer
f′(1) = 108
4.Derivatives
Answer: b = −16
Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.
Point and slope
f(2) = 12
f′(x) = 6x + 2 → f′(2) = 14
Tangent line at x = 0
b = f(x₀) − f′(x₀)·x₀ = 12 − 14·2 = −16
Answer
b = −16
5.Integrals
Answer: I = 3.75
Definite integrals: ∫ₐᵇ f(x) dx is the signed area between the graph and the x-axis from a to b. Find an antiderivative F (with F′ = f) and evaluate F(b) − F(a). Power rule: ∫xⁿ dx = xⁿ⁺¹/(n + 1). The area between two curves is ∫(upper − lower) dx.
Substitution
u = x² + 1, du = 2x dx
I = ∫12 u3 du
Integrate and evaluate
I = [u4/4]12 = (16 − 1)/4 = 3.75
Answer
I = 3.75
6.Integrals
Answer: V = 0.4189
Volumes of revolution: rotating the region under y = f(x) from x = a to x = b about the x-axis makes a solid whose cross-section at x is a disc of radius f(x). Adding the discs gives V = π∫ₐᵇ f(x)² dx. When the region lies between an outer curve R(x) and an inner curve r(x), each slice is a washer: V = π∫ₐᵇ (R(x)² − r(x)²) dx. Square the radii before you subtract them.
Intersection points
1x = 1x² → x = 0, x = 1
Washer method
V = π∫₀1 ((1x)² − (1x²)²) dx
Integrate
V = π(1·1³/3 − 1·1⁵/5) = π·0.1333 = 0.4189
Answer
V = 0.4189
7.Complex numbers
Answer: θ = −116.6 °
Polar form: z = r∠θ = r(cos θ + j sin θ) = r·ejθ, with r = |z| and θ = arg z. Back to rectangular form: a = r cos θ, b = r sin θ. The argument needs the quadrant: θ = atan(b/a) only when a > 0; for a < 0 add or subtract 180°. In polar form multiplication is easy: multiply the moduli and add the arguments; for division divide the moduli and subtract the arguments.
Quadrant
a = −1, b = −2 → III
Reference angle
α = atan(2/1) = 63.43°
Argument
θ = −(180° − α) = −116.6°
Answer
θ = −116.6°
In polar form z = 2.236∠−116.6°.
8.Complex numbers
Answer: a = 4
De Moivre and roots: (r∠θ)ⁿ = rⁿ∠nθ. The equation zⁿ = R∠φ has exactly n roots, all with modulus R1/n, at the arguments (φ + 360°·k)/n for k = 0, 1, …, n − 1: they lie evenly spaced on a circle, 360°/n apart. A quadratic z² + pz + q = 0 with p² < 4q has the complex roots z = −p/2 ± j√(q − p²/4); a polynomial with real coefficients always has its complex roots in conjugate pairs a ± jb.
Discriminant
p² − 4q = 64 − 80 = −16 < 0
Quadratic formula
z = (8 ± j√16)/2 = 4 + j2, 4 − j2
Answer
a = 4
Answer
a = 4
9.Differential equations
Answer: y(0.5) = 2.708
Second-order linear equations: for y″ + b·y′ + c·y = 0 try y = ert: this gives the characteristic equation r² + b·r + c = 0. Two real roots r₁ ≠ r₂ give y = A·er₁t + B·er₂t (overdamped); a double root r gives y = (A + Bt)·ert (critically damped); complex roots −α ± jω give y = e−αt(A cos ωt + B sin ωt) (underdamped, oscillating). The initial values y(0) and y′(0) fix A and B. Written as y″ + 2ζωny′ + ωn²y = 0, ωn is the natural frequency and ζ the damping ratio: ζ < 1 oscillates. With a constant input, y″ + by′ + cy = c·K, the solution is the steady value K plus the free response.
Characteristic roots
(r + 2)(r + 3) = 0 → r = −2, −3
Constants from the initial values
A + B = 5, −2A − 3B = -4
A = 11, B = −6
Evaluate
y(0.5) = 11·e−1 + (−6)·e−1.5 = 2.708
Answer
y(0.5) = 2.708
10.Differential equations
Answer: y(1.5) = 2.155
First-order differential equations: dy/dt = k·y has the solution y = y₀·ekt (growth for k > 0, decay for k < 0); a half-life T gives k = ln 2/T. Newton's law of cooling, dT/dt = −k·(T − Tₐ), gives T(t) = Tₐ + (T₀ − Tₐ)·e−kt. A linear equation y′ + p(x)·y = q(x) is solved with the integrating factor μ = e∫p dx: (μy)′ = μq, so μy = ∫μq dx + C.
Integrating factor
μ = e0.5t
(e0.5ty)′ = 4e−2.5t
General solution
y = 4/(−2.5)·e−3t + C·e−0.5t = −1.6·e−3t + C·e−0.5t
Initial condition
3 = −1.6 + C → C = 4.6
Evaluate
y(1.5) = −1.6·0.01111 + 4.6·0.4724 = 2.155
Answer
y(1.5) = 2.155
11.Matrices & determinants
Answer: det A = −30
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
2×2 minors of the first row
M₁₁ = 10, M₁₂ = −10, M₁₃ = 10
Cofactor expansion
det A = 1·10 − 0·(−10) + (−4)·10 = −30
Answer
det A = −30
12.Matrices & determinants
Answer: x = −0.9286
Matrices: the determinant of a 2×2 matrix is ad − bc. For 3×3, expand along a row: each entry times the determinant of the 2×2 matrix left after removing its row and column, with alternating signs + − +. Eigenvalues λ solve det(A − λI) = 0; for 2×2 that is λ² − (trace)·λ + det = 0.
Determinant of the coefficient matrix
det A = (−4)·2 − (−1)·(−6) = −14
Replace the x column
det Aₓ = 2·2 − (−1)·9 = 13
x = 13/(−14) = −0.9286
Answer
x = −0.9286
y = 1.714
13.Vectors: dot & cross product
Answer: compb a = −6.593
Dot product:a·b = axbx + ayby + azbz = |a|·|b|·cos θ, a scalar. It gives the angle between two vectors, cos θ = a·b/(|a|·|b|), and it is zero when they are perpendicular. The scalar projection of a onto b is a·b/|b|, and a constant force F moving through a displacement d does work W = F·d.
Dot product
a·b = (−4)·6 + 5·(−4) + (−4)·1 = −48
Length of b
|b| = √(6² + (−4)² + 1²) = 7.28
Scalar projection
compb a = a·b/|b| = −48/7.28 = −6.593
Answer
compb a = −6.593
Negative: a has a component pointing against b.
14.Vectors: dot & cross product
Answer: V = 104
Cross product:a × b = (aybz − azby, azbx − axbz, axby − aybx) is perpendicular to both vectors, and |a × b| = |a|·|b|·sin θ is the area of the parallelogram they span (the triangle has half of it). Order matters: b × a = −(a × b). The moment of a force F applied at point A about point O is MO = r × F, with r = A − O.
Cross product b × c
(b × c)x = bycz − bzcy = 0·(−2) − (−5)·(−5) = −25
(b × c)y = bzcx − bxcz = (−5)·(−1) − 3·(−2) = 11
(b × c)z = bxcy − bycx = 3·(−5) − 0·(−1) = −15
Dot product with a
a·(b × c) = 0·(−25) + (−4)·11 + 4·(−15) = −104
V = |−104| = 104
Answer
V = 104
15.Probability
Answer: P = 0.3077
Probability: with equally likely outcomes, P(A) = favourable/total. For independent events multiply: P(A and B) = P(A)·P(B). The complement often helps: P(at least one) = 1 − P(none). Binomial: in n independent trials with success probability p, P(X = k) = C(n, k)·pᵏ·(1 − p)ⁿ⁻ᵏ.
First and second draw
P = 8/14 · 7/13 = 56/182 = 0.3077
Answer
P = 0.3077
16.Probability
Answer: σY = 0.5356
Discrete random variables: a probability distribution lists every value x a variable can take with its probability p(x); the probabilities are between 0 and 1 and add up to 1. The expected value (mean) is E(X) = Σx·p(x), the long-run average. The variance is Var(X) = Σ(x − μ)²·p(x) = E(X²) − μ², and the standard deviation σ = √Var(X). For a linear change Y = aX + b: E(Y) = a·E(X) + b, Var(Y) = a²·Var(X) and σY = |a|·σX; adding a constant does not change the spread.
Spread of X
μ = 1.45, Var(X) = 1.147, σX = 1.071
Linear change
σY = |0.5|·1.071 = 0.5356
Answer
σY = 0.5356
The mean does move: E(Y) = 0.5·1.45 + 5 = 5.725.
17.Descriptive statistics
Answer: s = 17.19
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Descriptive statistics: the mean is the sum divided by n; the median is the middle value of the sorted data (the average of the two middle values when n is even). The sample standard deviation is s = √(Σ(x − x̄)²/(n − 1)). A z-score z = (x − μ)/σ says how many standard deviations a value lies from the mean.
Standardise
z = 0.6
Standard normal table
Φ(0.6) = 0.7257
P = 0.7257
Answer
P = 0.7257
19.Normal distribution
Answer: P = 0.9073
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.
Normal distribution: convert a value x to a z-score, z = (x − μ)/σ, which counts standard deviations from the mean. A z-table (or Φ(z)) gives the area to the left of z, which is P(X < x). The area to the right is 1 − Φ(z), and the area between two values is the difference of their two areas.